COMEDK 2024 Shift 2 Paper

To design the parallel plate capacitor with the given specifications, we need to determine the minimum area of the plates. Let's break down the given information and use it step by step.
The capacitor has a dielectric constant κ=5 and a dielectric strength of 106Vm−1. The voltage rating is 2 kV and the maximum electric field allowed is 10% of its dielectric strength.
First, calculate the maximum electric field:
Emax=0.1×106Vm−1=105Vm−1
Given the voltage rating, the separation between the plates (d) can be found using the relation:
V=Emax×d
Rearranging for d :
d=‌

V

Emax

=‌

2000V

105Vm−1

=0.02m
Next, we use the formula for the capacitance of a parallel plate capacitor:
C=κE0‌

A

d

Where:
‌C=60‌pF=60×10−12F
‌κ=5
‌E0=8.85×10−12Fm−1
‌d=0.02m
‌A=‌ Area of the plates ‌
Rearranging for A :
A=‌

C×d

κ×E0

Substituting the values:
Simplifying further, we get:
A≈2.71×10−2m2
Therefore, the minimum area of the plates should be:
Option B: 2.7×10−2m2